1. **State the problem:** Differentiate the implicit equation $$7y^2 + \sin(3x) = 12 - \frac{y}{4}$$ with respect to $$x$$. 2. **Rewrite the equation:** $$7y^2 + \sin(3x) = 12 - \frac{y}{4}$$. 3. **Differentiate both sides using implicit differentiation:** - Derivative of $$7y^2$$: $$7 \cdot 2y \cdot \frac{dy}{dx} = 14y \frac{dy}{dx}$$ by the chain rule. - Derivative of $$\sin(3x)$$: $$\cos(3x) \cdot 3 = 3 \cos(3x)$$ by the chain rule. - Derivative of constant 12: $$0$$. - Derivative of $$-\frac{y}{4}$$: $$-\frac{1}{4} \frac{dy}{dx}$$. 4. **Write the differentiated equation:** $$14y \frac{dy}{dx} + 3 \cos(3x) = - \frac{1}{4} \frac{dy}{dx}$$. 5. **Group terms with $$\frac{dy}{dx}$$ on the left:** $$14y \frac{dy}{dx} + \frac{1}{4} \frac{dy}{dx} = -3 \cos(3x)$$. 6. **Factor $$\frac{dy}{dx}$$:** $$\left(14y + \frac{1}{4}\right) \frac{dy}{dx} = -3 \cos(3x)$$. 7. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{-3 \cos(3x)}{14y + \frac{1}{4}} = \frac{-3 \cos(3x)}{14y + 0.25}$$. 8. **Final boxed answer:** $$\boxed{\frac{dy}{dx} = \frac{-3 \cos(3x)}{14y + \frac{1}{4}}}$$